Problem: There is a strip with an infinite number of cells. Cells are numbered starting with 0. Initially the cell i contains a ball with the number i.

There are n pockets located at cells a_1, …, a_n. Each cell contains at most one pocket.

Filtering is the following sequence of operations:

  * All pockets at cells a_1, …, a_n open simultaneously, which makes balls currently located at those cells disappear. After the balls disappear, the pockets close again.
  * For each cell i from 0 to ∞, if the cell i contains a ball, we move that ball to the free cell j with the lowest number. If there is no free cell j < i, the ball stays at the cell i.



Note that after each filtering operation each cell will still contain exactly one ball.

For example, let the cells 1, 3 and 4 contain pockets. The initial configuration of balls is shown below (underscores display the cells with pockets):

0 1 2 3 4 5 6 7 8 9 ... 

After opening and closing the pockets, balls 1, 3 and 4 disappear:

0  2  5 6 7 8 9 ... 

After moving all the balls to the left, the configuration looks like this:

0 2 5 6 7 8 9 10 11 12 ... 

Another filtering repetition results in the following:

0 5 8 9 10 11 12 13 14 15 ... 

You have to answer m questions. The i-th of these questions is "what is the number of the ball located at the cell x_i after k_i repetitions of the filtering operation?"

Input

The first line contains two integers n and m — the number of pockets and questions respectively (1 ≤ n, m ≤ 10^5).

The following line contains n integers a_1, …, a_n — the numbers of cells containing pockets (0 ≤ a_1 < … < a_n ≤ 10^9).

The following m lines describe questions. The i-th of these lines contains two integers x_i and k_i (0 ≤ x_i, k_i ≤ 10^9).

Output

Print m numbers — answers to the questions, in the same order as given in the input.

Example

Input

3 15
1 3 4
0 0
1 0
2 0
3 0
4 0
0 1
1 1
2 1
3 1
4 1
0 2
1 2
2 2
3 2
4 2


Output

0
1
2
3
4
0
2
5
6
7
0
5
8
9
10

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 100000;
int m, n;
int a[Maxn + 5], p[Maxn + 5];
int x[Maxn + 5], k[Maxn + 5];
int e_pos[Maxn + 5], steps[Maxn + 5];
int Root[Maxn + 5];
struct Segment_Node {
  int lson, rson;
  int val;
} seg[Maxn * 25 + 5];
int id_tot;
void build(int &root, int left = 1, int right = n) {
  root = ++id_tot;
  seg[root].val = right - left + 1;
  if (left == right) {
    return;
  }
  int mid = (left + right) >> 1;
  build(seg[root].lson, left, mid);
  build(seg[root].rson, mid + 1, right);
}
int find_kth(int root, int k, int left = 1, int right = n) {
  if (left == right) {
    return left;
  }
  int mid = (left + right) >> 1;
  if (k <= seg[seg[root].lson].val) {
    return find_kth(seg[root].lson, k, left, mid);
  }
  return find_kth(seg[root].rson, k - seg[seg[root].lson].val, mid + 1, right);
}
void update(int &root, int last, int x, int a, int left = 1, int right = n) {
  root = ++id_tot;
  seg[root] = seg[last];
  seg[root].val += a;
  if (left == right) {
    return;
  }
  int mid = (left + right) >> 1;
  if (x <= mid) {
    update(seg[root].lson, seg[last].lson, x, a, left, mid);
  } else {
    update(seg[root].rson, seg[last].rson, x, a, mid + 1, right);
  }
}
int query(int root, int x, int left = 1, int right = n) {
  if (left == right) {
    return seg[root].val;
  }
  int mid = (left + right) >> 1;
  if (x <= mid) {
    return query(seg[root].lson, x, left, mid);
  }
  return seg[seg[root].lson].val + query(seg[root].rson, x, mid + 1, right);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  e_pos[0] = a[1] - 1;
  steps[0] = 0;
  for (int i = 1; i < n; i++) {
    int delta = (a[i + 1] - e_pos[i - 1] - 1) / i;
    e_pos[i] = e_pos[i - 1] + i * delta;
    steps[i] = steps[i - 1] + delta;
  }
  for (int i = 2; i <= n; i++) {
    p[i] = (p[i - 1] + a[i] - a[i - 1]) % (i - 1);
  }
  build(Root[n]);
  for (int i = n; i > 0; i--) {
    p[i] = find_kth(Root[i], p[i] + 1);
    update(Root[i - 1], Root[i], p[i], -1);
  }
  for (int t = 1; t <= m; t++) {
    int x, k;
    scanf("%d%d", &x, &k);
    if (x < a[1]) {
      printf("%d\n", x);
      continue;
    }
    int bel = lower_bound(e_pos + 1, e_pos + n, x) - e_pos;
    k += steps[bel - 1] + (x - e_pos[bel - 1] - 1) / bel + 1;
    int ord = (x - e_pos[bel - 1] - 1) % bel + 1;
    ord = find_kth(Root[bel], ord);
    bel = lower_bound(steps + 1, steps + n, k) - steps;
    printf("%lld\n", e_pos[bel - 1] + 1ll * (k - steps[bel - 1] - 1) * bel +
                         query(Root[bel], ord));
  }
  return 0;
}