Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

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


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct node {
  long long int left, right, sum;
} tree[17][100005 * 4];
void update(int num, int root, int l, int r, int x, int val) {
  if (l == r) {
    tree[num][root].left = val;
    tree[num][root].right = val;
    tree[num][root].sum = val;
    return;
  }
  int m = (l + r) / 2;
  if (x <= m)
    update(num, root * 2, l, m, x, val);
  else
    update(num, root * 2 + 1, m + 1, r, x, val);
  if (tree[num][root * 2].left == m - l + 1)
    tree[num][root].left =
        tree[num][root * 2].left + tree[num][root * 2 + 1].left;
  else
    tree[num][root].left = tree[num][root * 2].left;
  if (tree[num][root * 2 + 1].right == r - m)
    tree[num][root].right =
        tree[num][root * 2].right + tree[num][root * 2 + 1].right;
  else
    tree[num][root].right = tree[num][root * 2 + 1].right;
  tree[num][root].sum =
      tree[num][root * 2].sum + tree[num][root * 2 + 1].sum +
      tree[num][root * 2].right * tree[num][root * 2 + 1].left;
}
int main() {
  int n, m, i, j, x, y;
  memset(tree, 0, sizeof(tree));
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    for (j = 0; j < 17; j++) update(j, 1, 1, n, i, bool(x & (1 << j)));
  }
  for (i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    for (j = 0; j < 17; j++) update(j, 1, 1, n, x, bool(y & (1 << j)));
    long long int ans = 0;
    for (j = 0; j < 17; j++) ans += tree[j][1].sum << j;
    printf("%lld\n", ans);
  }
}