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;
const int MX = 1e5 + 10, MD = 1e9 + 7, LG = 20;
long long n, m, a[MX], ans, tmp;
struct node {
  long long ans = 0, pre = 0, suf = 0;
} seg[LG][MX * 4];
node Merge(node a, node b, int e, int s) {
  int mid = (s + e) / 2;
  node res;
  res.ans = a.ans + b.ans - (a.suf * (a.suf + 1)) / 2 -
            (b.pre * (b.pre + 1)) / 2 +
            ((a.suf + b.pre) * (a.suf + b.pre + 1)) / 2;
  res.pre = a.pre;
  if (res.pre == mid - s) res.pre += b.pre;
  res.suf = b.suf;
  if (res.suf == e - mid) res.suf += a.suf;
  return res;
}
void build(int val, int id = 1, int s = 0, int e = n) {
  if (e - s == 1) {
    if (a[s] & (1 << val)) {
      seg[val][id].ans = 1;
      seg[val][id].suf = 1;
      seg[val][id].pre = 1;
    }
    return;
  }
  int mid = (s + e) / 2;
  build(val, id * 2, s, mid);
  build(val, id * 2 + 1, mid, e);
  seg[val][id] = Merge(seg[val][id * 2], seg[val][id * 2 + 1], e, s);
}
void update(int ind, int val2, int val, int id = 1, int s = 0, int e = n) {
  if (e - s == 1) {
    if (val == 1) {
      seg[val2][id].ans = 1;
      seg[val2][id].suf = 1;
      seg[val2][id].pre = 1;
    } else {
      seg[val2][id].ans = 0;
      seg[val2][id].suf = 0;
      seg[val2][id].pre = 0;
    }
    return;
  }
  int mid = (s + e) / 2;
  if (mid > ind)
    update(ind, val2, val, id * 2, s, mid);
  else
    update(ind, val2, val, id * 2 + 1, mid, e);
  seg[val2][id] = Merge(seg[val2][id * 2], seg[val2][id * 2 + 1], e, s);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < LG; i++) build(i);
  while (m--) {
    int p, v;
    cin >> p >> v;
    p--;
    ans = 0;
    for (int i = 0; i < LG; i++) {
      if (v & (1 << i)) {
        update(p, i, 1);
      } else {
        update(p, i, 0);
      }
      ans += seg[i][1].ans * (1 << i);
    }
    cout << ans << endl;
  }
}