#include <bits/stdc++.h>
using namespace std;

typedef pair <int, int> pii;

const int MAXN = 2e5 + 5;
const int offset = 1 << 18;
const int INF = 0x3f3f3f3f;

int N, Q;
vector <int> op[MAXN];
int bit[MAXN];
int dp[MAXN];
int suff[MAXN];
int t[2 * offset];

void load() {
	scanf("%d", &N);
	for (int i = 1; i <= N; i++)
		scanf("%d", bit + i);
	scanf("%d", &Q);
	while (Q--) {
		int l, r;
		scanf("%d%d", &l, &r);
		op[l].push_back(r);
	}
}

void update(int x, int lo, int hi, int from, int to, int val) {
	if (lo >= to || hi <= from)
		return;
	if (lo >= from && hi <= to) {
		t[x] = min(t[x], val);
		return;
	}
	int mid = (lo + hi) / 2;
	update(2 * x, lo, mid, from, to, val);
	update(2 * x + 1, mid, hi, from, to, val);
}

int query(int x) {
	int res = INF;
	for (x += offset; x; x /= 2)
		res = min(res, t[x]);
	return res;
}

int solve() {
	memset(t, INF, sizeof t);
	
	for (int i = N; i; i--)
		suff[i] = suff[i + 1] + !bit[i];
		
	for (int i = N; i; i--) {
		dp[i] = dp[i + 1] + bit[i];
		vector <pii> tmp;
		for (auto it : op[i]) {
			int val = dp[it + 1] - suff[it + 1];
			val = min(val, query(it));
			tmp.push_back({it, val});
			dp[i] = min(dp[i], val + suff[i]);
		}
		for (auto it : tmp)
			update(1, 0, offset, i, it.first, it.second);
	}
	
	return dp[1];
}

int main() {
	load();
	printf("%d\n", solve());
	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;

typedef pair <int, int> pii;

const int MAXN = 2e5 + 5;
const int offset = 1 << 18;
const int INF = 0x3f3f3f3f;

int N, Q;
vector <int> op[MAXN];
int bit[MAXN];
int dp[MAXN];
int suff[MAXN];
int t[2 * offset];

void load() {
	scanf("%d", &N);
	for (int i = 1; i <= N; i++)
		scanf("%d", bit + i);
	scanf("%d", &Q);
	while (Q--) {
		int l, r;
		scanf("%d%d", &l, &r);
		op[l].push_back(r);
	}
}

void update(int x, int lo, int hi, int from, int to, int val) {
	if (lo >= to || hi <= from)
		return;
	if (lo >= from && hi <= to) {
		t[x] = min(t[x], val);
		return;
	}
	int mid = (lo + hi) / 2;
	update(2 * x, lo, mid, from, to, val);
	update(2 * x + 1, mid, hi, from, to, val);
}

int query(int x) {
	int res = INF;
	for (x += offset; x; x /= 2)
		res = min(res, t[x]);
	return res;
}

int solve() {
	memset(t, INF, sizeof t);
	
	for (int i = N; i; i--)
		suff[i] = suff[i + 1] + !bit[i];
		
	for (int i = N; i; i--) {
		dp[i] = dp[i + 1] + bit[i];
		vector <pii> tmp;
		for (auto it : op[i]) {
			int val = dp[it + 1] - suff[it + 1];
			val = min(val, query(it));
			tmp.push_back({it, val});
			dp[i] = min(dp[i], val + suff[i]);
		}
		for (auto it : tmp)
			update(1, 0, offset, i, it.first, it.second);
	}
	
	return dp[1];
}

int main() {
	load();
	printf("%d\n", solve());
	return 0;
}
```