Problem: You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.

Output

Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value  - 1 if there is no such number.

Examples

Input

6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2


Output

2
6
-1
4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int fact[2 * 100005];
long long int func() {
  fact[0] = fact[1] = 1;
  for (long long int i = 2; i < 200001; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
  }
  return 0;
}
int arr[1000005];
int dp[1000005];
int main() {
  int n;
  int m;
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  dp[n] = n;
  for (int i = n - 1; i > 0; i--) {
    if (arr[i] == arr[i + 1]) {
      dp[i] = dp[i + 1];
    } else {
      dp[i] = i;
    }
  }
  int l, r, p;
  while (m--) {
    scanf("%d", &l);
    scanf("%d", &r);
    scanf("%d", &p);
    if (arr[l] != p) {
      printf("%d\n", l);
    } else {
      if (dp[l] >= r) {
        printf("-1\n");
      } else {
        printf("%d\n", dp[l] + 1);
      }
    }
  }
}