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 a[100005];
long long sum[20];
set<pair<int, int> > s[20];
long long ss(long long v) { return v * (v + 1) / 2; }
void solve() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d  ", &a[i]);
  }
  for (int bit = 0; bit < 20; ++bit) {
    for (int i = 1; i <= n;) {
      if (a[i] & (1 << bit)) {
        int j;
        for (j = i; j <= n; ++j) {
          if (a[j] & (1 << bit)) {
          } else {
            break;
          }
        }
        int len = j - i;
        s[bit].insert({i, j - 1});
        sum[bit] += ss(len);
        i = j;
      } else {
        ++i;
        continue;
      }
    }
  }
  for (int test = 0; test < m; ++test) {
    int ind, val;
    scanf("%d %d ", &ind, &val);
    long long ret = 0;
    for (int bit = 0; bit < 20; ++bit) {
      int g = (1 << bit);
      if ((a[ind] & g) == (val & g)) {
        ret += (long long)sum[bit] * (1 << bit);
        continue;
      }
      if (a[ind] & g) {
        set<pair<int, int> >::iterator it = s[bit].lower_bound({ind + 1, 0});
        it = prev(it);
        int l = (*it).first, r = (*it).second;
        int len = r - l + 1;
        sum[bit] -= ss(len);
        s[bit].erase(it);
        if (ind + 1 <= r) {
          s[bit].insert({ind + 1, r});
          sum[bit] += ss(r - ind - 1 + 1);
        }
        if (l <= ind - 1) {
          s[bit].insert({l, ind - 1});
          sum[bit] += ss(ind - 1 - l + 1);
        }
      } else {
        int l = ind, r = ind;
        set<pair<int, int> >::iterator x = s[bit].lower_bound({ind + 1, 0});
        auto it = s[bit].lower_bound({ind + 1, 0});
        if (it == s[bit].end() || (*it).first != ind + 1) {
        } else {
          int a = (*it).first, b = (*it).second;
          sum[bit] -= ss(b - a + 1);
          s[bit].erase(it);
          r = b;
        }
        auto jt = s[bit].lower_bound({ind + 1, 0});
        if (jt != s[bit].begin()) {
          jt = prev(jt);
          if ((*jt).second + 1 == ind) {
            int a = (*jt).first, b = (*jt).second;
            sum[bit] -= ss(b - a + 1);
            s[bit].erase(jt);
            l = a;
          }
        }
        sum[bit] += ss(r - l + 1);
        s[bit].insert({l, r});
      }
      ret += (long long)sum[bit] * (1 << bit);
    }
    a[ind] = val;
    printf("%lld\n", ret);
  }
}
int main() {
  solve();
  return 0;
}