Problem: Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:

* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.



Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.

Constraints

* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$

Input


$n$ $q$
$query_1$
$query_2$
:
$query_q$


In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:


0 $s$ $t$ $x$


or


1 $s$ $t$


The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.

Output

For each $find$ query, print the minimum value.

Example

Input

6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5


Output

-2
0
1
-1

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;

long long INF = (1 << 31) - 1;

class LazySegmentTree {
private:
	long long Size;
	vector<long long> Node, Lazy;
public:
	LazySegmentTree(long long N) {
		Size = 1;
		while (Size < N) Size <<= 1;
		Node.assign(Size * 2, 0);
		Lazy.assign(Size * 2, 0);
	}
	void Eval(long long K, long long L, long long R) {
		if (Lazy[K] != 0) {
			Node[K] += Lazy[K];
			if (R - L > 1) {
				Lazy[K * 2] += Lazy[K];
				Lazy[K * 2 + 1] += Lazy[K];
			}
			Lazy[K] = 0;
		}
	}
	void Update(long long A, long long B, long long X, long long K, long long L, long long R) {
		Eval(K, L, R);
		if (B <= L || R <= A) return;
		if (A <= L && R <= B) {
			Lazy[K] += X;
			Eval(K, L, R);
			return;
		}
		Update(A, B, X, K * 2, L, (L + R) / 2);
		Update(A, B, X, K * 2 + 1, (L + R) / 2, R);
		Node[K] = min(Node[K * 2], Node[K * 2 + 1]);
	}
	void Update(long long A, long long B, long long X) {
		Update(A, B, X, 1, 0, Size);
		return;
	}
	long long Query(long long A, long long B, long long K, long long L, long long R) {
		Eval(K, L, R);
		if (B <= L || R <= A) return INF;
		if (A <= L && R <= B) return Node[K];
		return min(Query(A, B, K * 2, L, (L + R) / 2), Query(A, B, K * 2 + 1, (L + R) / 2, R));
	}
	long long Query(long long A, long long B) {
		return Query(A, B, 1, 0, Size);
	}
};

int main() {
	long long N, Q;
	cin >> N >> Q;
	LazySegmentTree Seg(N);
	for (int i = 0; i < Q; i++) {
		long long Mode;
		cin >> Mode;
		if (Mode == 0) {
			long long A, B, X;
			cin >> A >> B >> X;
			Seg.Update(A, B + 1, X);
		}
		else {
			long long A, B;
			cin >> A >> B;
			cout << Seg.Query(A, B + 1) << endl;
		}
	}
}