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;
int n, m, a1[101000];
long long sum;
void find_limit(int tmp, long long& l, long long& r, int p) {
  l = r = 0;
  int i;
  for (i = p - 1; i; i--) {
    if ((a1[i] & tmp) == 0) break;
    l++;
  }
  for (i = p + 1; i <= n; i++) {
    if ((a1[i] & tmp) == 0) break;
    r++;
  }
}
void solve(int p, int new_value, int origin_value) {
  int i;
  long long cont_left = 0, cont_right = 0;
  for (i = 0; (1 << i) < 101000; i++) {
    int tmp = (1 << i);
    if ((new_value & tmp) == (origin_value & tmp)) continue;
    find_limit(tmp, cont_left, cont_right, p);
    long long sub_value =
        (cont_left * cont_right + cont_left + cont_right + 1) * (long long)tmp;
    if (new_value & tmp)
      sum += sub_value;
    else
      sum -= sub_value;
  }
}
void build() {
  sum = 0;
  int i, j;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &a1[i]);
  for (i = 0; (1 << i) < 101000; i++) {
    int tmp = (1 << i);
    long long num = 0;
    for (j = 1; j <= n; j++) {
      if (a1[j] & tmp)
        num++;
      else {
        if (num) sum += (num * (num + 1) * (long long)tmp / 2);
        num = 0;
      }
    }
    if (num) sum += (num * (num + 1) * (long long)tmp / 2);
  }
}
int main() {
  build();
  int p, v, i;
  for (i = 0; i < m; i++) {
    scanf("%d%d", &p, &v);
    solve(p, v, a1[p]);
    a1[p] = v;
    printf("%I64d\n", sum);
  }
  return 0;
}