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<cstdio>
#include<algorithm>
#define N 100000
#define MAX (1<<31)-1
using namespace std;
long long A[N]={0}, T[4*N], lazy[4*N]={0};

void build(int l, int r, int k) {
    if (l == r) {
        T[k] = A[l];
        return;
    }
    int mid = (l + r) / 2;
    build(l, mid, k * 2);
    build(mid + 1, r, k * 2 + 1);
    T[k] = min(T[2 * k], T[2 * k + 1]);
}
void pushdown(int k) {
    if (lazy[k]) {
        lazy[2*k] += lazy[k];
        lazy[2*k+1] += lazy[k];
        T[2*k] += lazy[k];
        T[2*k+1] += lazy[k];
        lazy[k] = 0;
    }
}
void updata(int L, int R, int data ,int l, int r ,int k) {
    if (L <= l && r <= R) {
        T[k] += data;
        lazy[k] += data;
        return;
    }
    int mid = (l + r) / 2;
    pushdown(k);
    if (mid >= L) updata(L, R, data, l, mid, 2*k);
    if (mid < R) updata(L, R, data, mid+1, r, 2*k+1);
    T[k] = min(T[2 * k], T[2 * k + 1]);
}
long long find(int L, int R, int l, int r, int k) {
    if (L <= l && r <= R) return T[k];
    int mid = (l + r) / 2;
    pushdown(k);
    long long ans = MAX;
    if (mid >= L) ans = min(ans, find(L, R, l, mid, k * 2));
    if (mid < R) ans = min(ans, find(L, R, mid + 1, r, k * 2 + 1));
    return ans;
}
int main() {
    int n, q;
    scanf("%d%d", &n, &q);
    build(0, n-1, 1);
    int com, s, t, x, y;
    while (q--) {
        scanf("%d", &com);
        if (com == 0) {
            scanf("%d%d%d", &s, &t, &x);
            updata(s, t, x, 0, n-1, 1);
        }
        else {
            scanf("%d%d", &s, &t);
            printf("%ld\n", find(s, t, 0, n-1, 1));
        }
    }
    return 0;
}