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 N = 100005;
const int M = 20;
int n, m, a[N];
set<pair<int, int> > s[M];
long long ans = 0;
long long sum(int a, int b) { return 1LL * a * (a + 1) / 2 * (1LL << b); }
void insert(set<pair<int, int> > &s, int p, int bit) {
  pair<int, int> now = make_pair(p, p);
  set<pair<int, int> >::iterator it = s.lower_bound(make_pair(p, p));
  if (it != s.begin()) {
    it--;
    pair<int, int> l = *(it);
    if (l.second == now.first - 1) {
      now.first = l.first;
      s.erase(it);
      ans -= sum(l.second - l.first + 1, bit);
    }
  }
  it = s.lower_bound(make_pair(p, p));
  if (it != s.end()) {
    pair<int, int> r = *(it);
    if (r.first == now.second + 1) {
      now.second = r.second;
      s.erase(it);
      ans -= sum(r.second - r.first + 1, bit);
    }
  }
  ans += sum(now.second - now.first + 1, bit);
  s.insert(now);
}
void erase(set<pair<int, int> > &s, int p, int bit) {
  set<pair<int, int> >::iterator it = s.upper_bound(make_pair(p, n + 1));
  it--;
  pair<int, int> now = *it;
  ans -= sum(now.second - now.first + 1, bit);
  s.erase(it);
  if (p != now.first) {
    s.insert(make_pair(now.first, p - 1));
    ans += sum(p - 1 - now.first + 1, bit);
  }
  if (p != now.second) {
    s.insert(make_pair(p + 1, now.second));
    ans += sum(now.second - (p + 1) + 1, bit);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < M; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[j] & (1 << i)) {
        insert(s[i], j, i);
      }
    }
  }
  while (m--) {
    int p, v;
    scanf("%d %d", &p, &v);
    for (int i = 0; i < M; i++) {
      if (v & (1 << i)) {
        if (!(a[p] & (1 << i))) insert(s[i], p, i);
      } else {
        if (a[p] & (1 << i)) erase(s[i], p, i);
      }
    }
    a[p] = v;
    printf("%I64d\n", ans);
  }
  return 0;
}