#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const ll INF = 1LL << 60;
class BiOp {
public:
	ll operator()(ll a, ll b) const {
		return min(a, b);
	}
};

template<typename T, typename BiOp>
struct SegmentTree {
	vector<T> data;
	T e;
	BiOp op;
	int size;

	SegmentTree(int n, T e, BiOp op) : e(e), op(op) {
		size = 1;
		while (size < n) size *= 2;
		data.resize(2 * size - 1);
		for (int i = 0; i < 2 * size - 1; i++) data[i] = e;
	}

	void update(int k, T v) {
		k += size - 1;
		data[k] = v;
		while (k > 0) {
			k = (k - 1) / 2;
			data[k] = op(data[k * 2 + 1], data[k * 2 + 2]);
		}
	}

	T query(int a, int b, int k, int l, int r) const {
		if (b <= l || r <= a) return e;
		if (a <= l && r <= b) return data[k];
		return op(query(a, b, k * 2 + 1, l, (l + r) / 2), query(a, b, k * 2 + 2, (l + r) / 2, r));
	}

	T query(int a, int b) {
		return query(a, b, 0, 0, size);
	}

	T get(int i) {
		return data[i + size - 1];
	}
};
int b[200000];
vector<int> R[200000];
int main() {
	int N;
	cin >> N;
	int zero = 0;
	for (int i = 0; i < N; i++) {
		cin >> b[i];
		if (b[i] == 0) zero++;
	}
	int Q;
	cin >> Q;
	for (int i = 0; i < Q; i++) {
		int l, r;
		cin >> l >> r;
		l--; r--;
		R[l].push_back(r);
	}
	SegmentTree<ll, BiOp> seg(N + 1, INF, BiOp());
	for (int r : R[0]) seg.update(r, 0);
	ll x = (b[0] == 1) ? 1 : -1;
	seg.update(0, min(seg.get(0), x));
	for (int i = 1; i < N; i++) {
		for (int r : R[i]) {
			ll m = seg.query(i - 1, r + 1);
			seg.update(r, m);
		}
		x = (b[i] == 1) ? 1 : -1;
		seg.update(i, min(seg.get(i - 1) + x, seg.get(i)));
	}
	cout << zero + seg.get(N - 1) << endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 long long ll;
const ll INF = 1LL << 60;
class BiOp {
public:
	ll operator()(ll a, ll b) const {
		return min(a, b);
	}
};

template<typename T, typename BiOp>
struct SegmentTree {
	vector<T> data;
	T e;
	BiOp op;
	int size;

	SegmentTree(int n, T e, BiOp op) : e(e), op(op) {
		size = 1;
		while (size < n) size *= 2;
		data.resize(2 * size - 1);
		for (int i = 0; i < 2 * size - 1; i++) data[i] = e;
	}

	void update(int k, T v) {
		k += size - 1;
		data[k] = v;
		while (k > 0) {
			k = (k - 1) / 2;
			data[k] = op(data[k * 2 + 1], data[k * 2 + 2]);
		}
	}

	T query(int a, int b, int k, int l, int r) const {
		if (b <= l || r <= a) return e;
		if (a <= l && r <= b) return data[k];
		return op(query(a, b, k * 2 + 1, l, (l + r) / 2), query(a, b, k * 2 + 2, (l + r) / 2, r));
	}

	T query(int a, int b) {
		return query(a, b, 0, 0, size);
	}

	T get(int i) {
		return data[i + size - 1];
	}
};
int b[200000];
vector<int> R[200000];
int main() {
	int N;
	cin >> N;
	int zero = 0;
	for (int i = 0; i < N; i++) {
		cin >> b[i];
		if (b[i] == 0) zero++;
	}
	int Q;
	cin >> Q;
	for (int i = 0; i < Q; i++) {
		int l, r;
		cin >> l >> r;
		l--; r--;
		R[l].push_back(r);
	}
	SegmentTree<ll, BiOp> seg(N + 1, INF, BiOp());
	for (int r : R[0]) seg.update(r, 0);
	ll x = (b[0] == 1) ? 1 : -1;
	seg.update(0, min(seg.get(0), x));
	for (int i = 1; i < N; i++) {
		for (int r : R[i]) {
			ll m = seg.query(i - 1, r + 1);
			seg.update(r, m);
		}
		x = (b[i] == 1) ? 1 : -1;
		seg.update(i, min(seg.get(i - 1) + x, seg.get(i)));
	}
	cout << zero + seg.get(N - 1) << endl;
}
```