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#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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#define int long long
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#define endl "\n"
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/*
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http://ybt.ssoier.cn:8088/problem_show.php?pid=1642
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测试用例：
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5 1000
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答案：
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5
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*/
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const int N = 2;
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int n, mod;
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// 矩阵乘法
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void mul(int a[][N], int b[][N], int c[][N]) {
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int t[N][N] = {0};
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for (int i = 0; i < N; i++) {
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for (int j = 0; j < N; j++)
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for (int k = 0; k < N; k++)
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t[i][j] = (t[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
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}
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memcpy(c, t, sizeof t);
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}
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signed main() {
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cin >> n >> mod;
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int b[N][N] = {1, 1}; // 结果矩阵(初始化fib[2]=1,fib[1]=1)
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// 构造的向量数组
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int m[N][N] = {
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{1, 1},
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{1, 0}};
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// 矩阵快速幂
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for (int i = n - 1; i; i >>= 1) {
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if (i & 1) mul(b, m, b);
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mul(m, m, m);
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}
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printf("%lld\n", b[0][1]);
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return 0;
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}
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