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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e17;

template<typename T, typename E, typename F, typename G, typename H>
class LazySegmentTree {
  int n, height;
  F f; G g; H h; const T ti; const E ei;
  vector<T> dat; vector<E> laz; vector<int> highest;

  void propagate(int k) {
    if(laz[k] != ei) {
      laz[2*k] = h(laz[2*k],laz[k]); laz[2*k+1] = h(laz[2*k+1],laz[k]);
      dat[k] = g(dat[k],laz[k],n/highest[k]);
      laz[k] = ei;
    }
  }
  inline void recalc(int k) {
    while(k >>= 1) dat[k] = f(g(dat[2*k],laz[2*k],n/highest[2*k]),g(dat[2*k+1],laz[2*k+1],n/highest[2*k+1]));
  }
  inline void thrust(int k) {
    for(int i = height; i > 0; i--) propagate(k >> i);
  }
public:
  LazySegmentTree(int _n, F f, G g, H h, T ti, E ei): f(f), g(g), h(h), ti(ti), ei(ei) {
    n = 1; height = 0;
    while(n < _n) n <<= 1, height++;
    dat.assign(2*n,ti); laz.assign(2*n,ei); highest.assign(2*n,0);
    int c = 1;
    for(int i = 1; i < 2*n; i++) {
      if(c*2 <= i) c <<= 1;
      highest[i] = c;
    }
  }
  void set(int k, T x) {
    dat[k+n] = x;
  }
  void build() {
    for(int i = n-1; i >= 0; i--) {
      dat[i] = f(dat[2*i],dat[2*i+1]);
    }
  }
  void update(int a, int b, const E x) {
    thrust(a += n); thrust(b += n-1);
    for(int l = a, r = b + 1; l < r; l >>= 1, r >>= 1) {
      if(l & 1) laz[l] = h(laz[l],x), ++l;
      if(r & 1) --r, laz[r] = h(laz[r],x);
    }
    recalc(a); recalc(b);
  }
  T query(int a, int b) {
    thrust(a += n); thrust(b += n-1);
    T l = ti, r = ti; b++;
    while(a < b) {
      if(a & 1) l = f(l,g(dat[a],laz[a],n/highest[a])), a++;
      if(b & 1) b--, r = f(g(dat[b],laz[b],n/highest[b]),r);
      a >>= 1; b >>= 1;
    }
    return f(l,r);
  }
};

int main(){
  int n, q;
  cin >> n >> q;
  auto f = [](ll a, ll b){return min(a,b);};
  auto g = [](ll a, ll b, int l){return a+b;};
  auto h = [](ll a, ll b){return a+b;};
  LazySegmentTree<ll,ll,decltype(f),decltype(g),decltype(h)> lst(n+1,f,g,h,INF,0);
  rep(i,n) lst.set(i,0);
  lst.build();
  vector<ll> ans;
  rep(i,q) {
    int t;
    cin >> t;
    if(t == 0) {
      int l, r; ll x;
      cin >> l >> r >> x;
      lst.update(l,r+1,x);
    } else {
      int l, r;
      cin >> l >> r;
      ans.push_back(lst.query(l,r+1));
    }
  }
  for(auto x : ans) cout << x << endl;
  return 0;
}