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 maxn = 1e5 + 20;
const int maxb = 19;
struct node {
  int prefix, suffix, size;
  long long sum;
  node() { prefix = suffix = size = sum = 0; }
};
node seg[maxn * 4][maxb];
int a[maxn];
node merge(node a, node b) {
  node c;
  c.prefix = ((a.prefix == a.size) ? a.size + b.prefix : a.prefix);
  c.suffix = ((b.suffix == b.size) ? b.size + a.suffix : b.suffix);
  c.sum = a.sum + b.sum + (long long)a.suffix * b.prefix;
  c.size = a.size + b.size;
  return c;
}
void build(int s, int e, int v, int b) {
  if (e - s < 2) {
    seg[v][b].size = 1;
    seg[v][b].prefix = seg[v][b].suffix = seg[v][b].sum = (((a[s]) >> (b)) & 1);
    return;
  }
  int m = (s + e) / 2;
  build(s, m, 2 * v, b);
  build(m, e, 2 * v + 1, b);
  seg[v][b] = merge(seg[2 * v][b], seg[2 * v + 1][b]);
}
void update(int p, int s, int e, int v, int b) {
  if (e - s < 2) {
    seg[v][b].size = 1;
    seg[v][b].prefix = seg[v][b].suffix = seg[v][b].sum = (((a[s]) >> (b)) & 1);
    return;
  }
  int m = (s + e) / 2;
  if (p < m)
    update(p, s, m, 2 * v, b);
  else
    update(p, m, e, 2 * v + 1, b);
  seg[v][b] = merge(seg[2 * v][b], seg[2 * v + 1][b]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < maxb; i++) build(0, n, 1, i);
  while (q--) {
    int p;
    cin >> p;
    p--;
    cin >> a[p];
    for (int i = 0; i < maxb; i++) update(p, 0, n, 1, i);
    long long ans = 0;
    for (int i = 0; i < maxb; i++) ans += (1LL << i) * seg[1][i].sum;
    cout << ans << endl;
  }
}