#include <bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;
#define int long long
#define endl "\n"
/*
http://ybt.ssoier.cn:8088/problem_show.php?pid=1642
测试用例：
5 1000

答案：
5
*/
const int N = 2;

int n, mod;

// 矩阵乘法
void mul(int a[][N], int b[][N], int c[][N]) {
    int t[N][N] = {0};
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++)
            for (int k = 0; k < N; k++)
                t[i][j] = (t[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
    }
    memcpy(c, t, sizeof t);
}

signed main() {
    cin >> n >> mod;

    int b[N][N] = {1, 1}; // 结果矩阵(初始化fib[2]=1,fib[1]=1)

    // 构造的向量数组
    int m[N][N] = {
        {1, 1},
        {1, 0}};

    // 矩阵快速幂
    for (int i = n - 1; i; i >>= 1) {
        if (i & 1) mul(b, m, b);
        mul(m, m, m);
    }

    printf("%lld\n", b[0][1]);
    return 0;
}