设 x 为一次查询的区间长度（即 x = R - L + 1），则使得 lg[x] = 5 的 x 的取值范围是（ ）。

#include <iostream>
using namespace std;
int n, m, a[100007], L, R, lg[100007], i, j, t, dp[100007][25], pw[25];
int gcd(int x, int y) {
    if (y == 0) return x;
    return gcd(y, x % y);
}
int main() {
    cin >> n >> m;
    for (i = 1; i <= n; i++) cin >> a[i];
    t = 0;
    pw[0] = 1;
    for (i = 1; i <= 24; i++) pw[i] = pw[i - 1] * 2;
    for (i = 2; i <= 100000; i++)
        if (pw[t + 1] >= i) lg[i] = t;
        else t++, lg[i] = t;
    for (i = 1; i <= n; i++)
        dp[i][0] = a[i];
    for (j = 1; j <= lg[n]; j++)
        for (i = 1; i + pw[j] - 1 <= n; i++) {
            dp[i][j] = gcd(dp[i][j - 1], dp[i + pw[j] - 1][j - 1]);
        }
    for (i = 1; i <= m; i++) {
        cin >> L >> R;
        cout << gcd(dp[L][lg[R-L+1]],dp[R-pw[lg[R-L+1]]+1][lg[R-L+1]]) << endl;
    }
    return 0;
}