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;
vector<pair<pair<int, int>, int> > query;
vector<int> open[maxn], close[maxn];
set<pair<int, int> > st;
int ans[maxn];
int MAIN() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    int d;
    scanf("%d", &d);
    a.push_back(d);
  }
  for (int i = 0; i < m; ++i) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    --l;
    query.push_back(make_pair(make_pair(l, r), x));
    open[l].push_back(i), close[r].push_back(i);
  }
  memset(ans, -1, sizeof(ans));
  for (int i = 0; i < n; ++i) {
    vector<int>::iterator it;
    for (it = close[i].begin(); it != close[i].end(); ++it) {
      int id = *it;
      if (ans[id] == -1) {
        ans[id] = -2;
        st.erase(make_pair(query[id].second, id));
      }
    }
    for (it = open[i].begin(); it != open[i].end(); ++it) {
      int id = *it;
      st.insert(make_pair(query[id].second, id));
    }
    int x = a[i];
    set<pair<int, int> >::iterator sit, sit2;
    sit = st.lower_bound(make_pair(x + 1, -1));
    while (sit != st.end()) {
      ans[sit->second] = i;
      st.erase(sit++);
    }
    sit = st.lower_bound(make_pair(x, -1));
    if (sit != st.begin()) {
      --sit;
      while (true) {
        ans[sit->second] = i;
        if (sit == st.begin()) {
          st.erase(sit);
          break;
        } else {
          st.erase(sit--);
        }
      }
    }
  }
  vector<int>::iterator it;
  for (it = close[n].begin(); it != close[n].end(); ++it) {
    int id = *it;
    if (ans[id] == -1) {
      ans[id] = -2;
      st.erase(make_pair(query[id].second, id));
    }
  }
  for (int i = 0; i < m; ++i) {
    printf("%d\n", ans[i] + 1);
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int ret = MAIN();
  return ret;
}