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#include <bits/stdc++.h>
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const int N = 510;
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//性能：第12号测试点，时间最长，65ms
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/*
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搜索代码：暴力枚举每次选择哪个点，能选就选，维护剩下几个自由点，加个记忆化即可通过
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*/
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using namespace std;
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int n, m, ans;
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int x[N], y[N]; //对于二维坐标，两个x,y数组
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int f[N][N]; //结果数组
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//从u点出发，还有r个虚拟点可用,可以获得的最长序列长度是多少
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int dfs(int u, int r) {
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if (~f[u][r]) return f[u][r]; //计算过则直接返回
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int ans = r + 1; //剩余r个虚拟点，再加上当前点u，最起码能有r+1点的序列长度
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for (int i = 1; i <= n; i++) { //谁能做为我的后续点
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if (u == i) continue; //自己不能做为自己的直接后续点
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if (x[i] < x[u] || y[i] < y[u]) continue; //排除掉肯定不可能成为我后续点的点
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int d = x[i] - x[u] + y[i] - y[u] - 1; // u->i之间缺少 多少个虚拟点
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if (d > r) continue; //如果需要的虚拟点个数大于剩余的虚拟点个数，那么i 无法做为u的后续点
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ans = max(ans, dfs(i, r - d) + d + 1); //在消耗了d个虚拟点之后，成功到达了i这个点；
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//①已经取得的序列长度贡献u和d个虚拟点，共d+1个
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//②问题转化为求未知部分:以i点为出发点，剩余虚拟点个数r-d个的情况下可以获取到的最长序列长度
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}
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return f[u][r] = ans; //记忆化
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}
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int main() {
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//文件输入
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// freopen("point.in", "r", stdin);
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memset(f, -1, sizeof f);
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cin >> n >> m;
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for (int i = 1; i <= n; i++) cin >> x[i] >> y[i];
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for (int i = 1; i <= n; i++) ans = max(ans, dfs(i, m));
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printf("%d\n", ans);
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return 0;
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}
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