#include <bits/stdc++.h>
#define mk make_pair
#define ll long long
using namespace std;
const int N = 2e5 + 5;
int n, q, a[N], s0[N], s1[N], ans[N];
pair<int, int> b[N];
int c[N * 4]; 
void build(int k, int l, int r) {
	if (l == r) {
		c[k] = 1e9;
		if (l == 0) c[k] = 0;
		return;
	}
	int mid = l + r >> 1;
	build(k << 1, l, mid);
	build(k << 1 | 1, mid + 1, r);
	c[k] = min(c[k << 1], c[k << 1 | 1]);
}
int Query(int k, int l, int r, int p, int q) {
	if (l >= p && r <= q) return c[k];
	if (l > q || r < p) return 1e9;
	int mid = l + r >> 1;
	return min(Query(k << 1, l, mid, p, q), Query(k << 1 | 1, mid + 1, r, p, q));
}
void modify(int k, int l, int r, int p, int v) {
	if (l == r) {
		c[k] = min(c[k], v);
		return;
	}
	int mid = l + r >> 1;
	if (p <= mid) 
		modify(k << 1, l, mid, p, v);
	else 
		modify(k << 1 | 1, mid + 1, r, p, v);
	c[k] = min(c[k << 1], c[k << 1 | 1]);
}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		scanf("%d", a + i);
		s0[i] = s0[i - 1] + (a[i] == 0);
		s1[i] = s1[i - 1] + (a[i] == 1);
	}
	cin >> q;
	for (int i = 1; i <= q; i++) 
		scanf("%d %d", &b[i].first, &b[i].second);
	sort(b + 1, b + 1 + q);
	for (int i = 1; i <= n; i++) ans[i] = 1e9;
	build(1, 0, n);
	int p = 1;
	for (int i = 1; i <= n; i++) {
		while (p <= q && b[p].first == i) {
			int tmp = Query(1, 0, n, b[p].first - 1, b[p].second);
			tmp += s0[b[p].second];
			tmp -= s1[b[p].second];
			ans[b[p].second] = min(ans[b[p].second], tmp);
			modify(1, 0, n, b[p].second, tmp - s0[b[p].second] + s1[b[p].second]); 
			p++;
		}
		ans[i] = min(ans[i], ans[i - 1]);
		modify(1, 0, n, i, ans[i] - s0[i] + s1[i]);
	}
	cout << ans[n] + s1[n] << endl;
}

### Prompt

Generate a Cpp solution to the following problem:

You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

### Response

```cpp
#include <bits/stdc++.h>
#define mk make_pair
#define ll long long
using namespace std;
const int N = 2e5 + 5;
int n, q, a[N], s0[N], s1[N], ans[N];
pair<int, int> b[N];
int c[N * 4]; 
void build(int k, int l, int r) {
	if (l == r) {
		c[k] = 1e9;
		if (l == 0) c[k] = 0;
		return;
	}
	int mid = l + r >> 1;
	build(k << 1, l, mid);
	build(k << 1 | 1, mid + 1, r);
	c[k] = min(c[k << 1], c[k << 1 | 1]);
}
int Query(int k, int l, int r, int p, int q) {
	if (l >= p && r <= q) return c[k];
	if (l > q || r < p) return 1e9;
	int mid = l + r >> 1;
	return min(Query(k << 1, l, mid, p, q), Query(k << 1 | 1, mid + 1, r, p, q));
}
void modify(int k, int l, int r, int p, int v) {
	if (l == r) {
		c[k] = min(c[k], v);
		return;
	}
	int mid = l + r >> 1;
	if (p <= mid) 
		modify(k << 1, l, mid, p, v);
	else 
		modify(k << 1 | 1, mid + 1, r, p, v);
	c[k] = min(c[k << 1], c[k << 1 | 1]);
}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		scanf("%d", a + i);
		s0[i] = s0[i - 1] + (a[i] == 0);
		s1[i] = s1[i - 1] + (a[i] == 1);
	}
	cin >> q;
	for (int i = 1; i <= q; i++) 
		scanf("%d %d", &b[i].first, &b[i].second);
	sort(b + 1, b + 1 + q);
	for (int i = 1; i <= n; i++) ans[i] = 1e9;
	build(1, 0, n);
	int p = 1;
	for (int i = 1; i <= n; i++) {
		while (p <= q && b[p].first == i) {
			int tmp = Query(1, 0, n, b[p].first - 1, b[p].second);
			tmp += s0[b[p].second];
			tmp -= s1[b[p].second];
			ans[b[p].second] = min(ans[b[p].second], tmp);
			modify(1, 0, n, b[p].second, tmp - s0[b[p].second] + s1[b[p].second]); 
			p++;
		}
		ans[i] = min(ans[i], ans[i - 1]);
		modify(1, 0, n, i, ans[i] - s0[i] + s1[i]);
	}
	cout << ans[n] + s1[n] << endl;
}

```