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;
const long long INF = 4e18 + 5;
const int inf = 1e9 + 5;
const int maxn = 2 * 1e5 + 5;
int n, m;
vector<int> a;
int f[maxn];
int MAIN() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    int d;
    scanf("%d", &d);
    a.push_back(d);
  }
  f[0] = -inf;
  for (int i = 1; i < n; ++i) {
    f[i] = (a[i] == a[i - 1] ? f[i - 1] : i - 1);
  }
  for (int i = 0; i < m; ++i) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    --l, --r;
    if (a[r] != x)
      printf("%d\n", r + 1);
    else {
      if (l <= f[r])
        printf("%d\n", f[r] + 1);
      else
        puts("-1");
    }
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int ret = MAIN();
  return ret;
}