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>
int a[16 + 1][100000], n, m, i, j, c, x;
unsigned long long s, k, kl, kr, t;
int main() {
  scanf("%d%d", &n, &m);
  for (k = 0; k < n; k++) {
    scanf("%d", &c);
    for (i = 0; i <= 16; i++)
      if (c & 1 << i) a[i][k] = 1;
  }
  for (i = 0; i <= 16; i++) {
    k = 0;
    for (j = 0; j < n; j++)
      if (a[i][j])
        ++k;
      else {
        if (k) s += (unsigned long long)(1ULL << i) * (k * (k + 1) / 2);
        k = 0;
      }
    if (k) s += (unsigned long long)(1ULL << i) * (k * (k + 1) / 2);
  }
  while (m--) {
    scanf("%d%d", &x, &c);
    --x;
    for (i = 0; i <= 16; i++)
      if (!(c & 1 << i) == a[i][x]) {
        kl = kr = 0;
        t = a[i][x] ? -1 : 1;
        a[i][x] += t;
        for (j = x - 1; j >= 0 && a[i][j]; j--) ++kl;
        for (j = x + 1; j < n && a[i][j]; j++) ++kr;
        s += (unsigned long long)t * (1ULL << i) *
             ((kl + kr + 1) * (kl + kr + 2) / 2 - kl * (kl + 1) / 2 -
              kr * (kr + 1) / 2);
      }
    printf("%I64u\n", s);
  }
  return 0;
}