Problem: Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries: 

  1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x. 
  2. Second type has form 2 y and asks you to find value of GukiZiana(a, y). 



For each query of type 2, print the answer and make GukiZ happy!

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries. 

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a. 

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.

Output

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

Examples

Input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3


Output

2


Input

2 3
1 2
1 2 2 1
2 3
2 4


Output

0
-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
long long add[750], a[500125];
int SZ;
vector<pair<long long, int>> block[750];
void rset(int i) {
  int st = i * SZ, ed = min(n, st + SZ);
  block[i].clear();
  for (int j = st; j < ed; j++) {
    a[j] += add[i];
    block[i].push_back({a[j], j});
  }
  add[i] = 0;
  sort(block[i].begin(), block[i].end());
}
void add1(int l, int r, int x) {
  int lid = l / SZ, rid = r / SZ;
  if (lid == rid) {
    for (int i = l; i <= r; i++) a[i] += x;
    rset(lid);
    return;
  }
  for (int i = l; i < lid * SZ + SZ; i++) a[i] += x;
  rset(lid);
  for (int j = lid + 1; j < rid; j++) add[j] += x;
  for (int i = rid * SZ; i <= r; i++) a[i] += x;
  rset(rid);
}
int qy(long long x) {
  int l = -1, r;
  for (int id = 0; id * SZ < n; id++) {
    long long tar = x - add[id];
    auto it =
        lower_bound(block[id].begin(), block[id].end(), make_pair(tar, -1));
    if (it == block[id].end()) continue;
    if ((*it).first == tar) {
      l = (*it).second;
      break;
    }
  }
  if (l == -1) return -1;
  for (int id = n / SZ; id >= 0; id--) {
    if (block[id].empty()) continue;
    long long tar = x - add[id];
    auto it = upper_bound(block[id].begin(), block[id].end(),
                          make_pair(tar, (1 << 30)));
    if (it == block[id].begin()) continue;
    it--;
    if ((*it).first == tar) {
      r = (*it).second;
      break;
    }
  }
  return r - l;
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    scanf("%lld", a + i);
  }
  SZ = sqrt(n);
  for (int i = 0; i * SZ < n; i++) rset(i);
  for (int t, l, r, x, i = 0; i < q; i++) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &l, &r, &x);
      l--;
      r--;
      add1(l, r, x);
    } else {
      scanf("%d", &x);
      printf("%d\n", qy(x));
    }
  }
  return 0;
}