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// https://www.jisuanke.com/problem/T3697
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010;
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int a[N], b[N], c[N];
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/*
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测试用例：
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5 5 0 1
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1 2 3 4 5
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1 2 3 4 5
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n=5 首歌
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m=5 喜欢的歌长度小于5
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k=0 没有不喜欢的
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p=1 满意值第一大的是哪首歌
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输入每首歌的长度a[i]
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输入每首歌的喜欢程度值b[i]
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答案：
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5
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*/
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struct Node {
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int id;
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int value;
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const bool operator<(const Node &b) const {
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return value > b.value;
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}
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} d[N];
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int dl;
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int main() {
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// 必须符合计蒜客的输入输出要求，否则不能完成提交！
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freopen("song.in", "r", stdin);
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freopen("song.out", "w", stdout);
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// n首歌，m:喜欢歌的长度a[i]<m,k：不喜欢歌的数量,p:满意值第p大
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int n, m, k, p;
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cin >> n >> m >> k >> p;
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for (int i = 1; i <= n; i++) cin >> a[i]; // 歌的长度
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for (int i = 1; i <= n; i++) cin >> b[i]; // 满意值
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// 超过长度m限制不喜欢
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for (int i = 1; i <= n; i++)
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if (a[i] > m) c[i] = 1;
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// 特殊指定的不喜欢
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for (int i = 1; i <= k; i++) {
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int x;
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cin >> x;
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c[x] = 1; // 不喜欢,当桶用
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}
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for (int i = 1; i <= n; i++)
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if (!c[i]) d[dl++] = {i, b[i]};
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sort(d, d + dl);
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if (d[p - 1].value)
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printf("%d\n", d[p - 1].id);
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else
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puts("aaaaaaaaa");
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return 0;
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}
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