This problem can be solved by checking if \(A _ 1+A _ 2+A _ 3+\cdots+A _ N\) is larger than \(M\).

One can compute \(A _ 1+A _ 2+A _ 3+\cdots+A _ N\) using a for statement, and compare it with \(M\) using an if statement.

The following is sample code.

#include <iostream>
using namespace std;

int main() {
    int N, M;
    cin >> N >> M;

    // Find the sum of sizes
    int sum = 0;
    for (int i = 0; i < N; ++i) {
        int A;
        cin >> A;
        sum += A;
    }

    // Print Yes if the sum is M or less
    if (sum <= M) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
    return 0;
}

N, M = map(int, input().split())

if sum(map(int, input().split())) <= M:
    print('Yes')
else:
    print('No')