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
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

// 遅延セグメントツリー
template <typename T, typename E>
struct segtree {
  using F = function<T(T,T)>;
  using G = function<T(T,E)>;
  using H = function<E(E,E)>;
  using P = function<E(E,int)>;
  F f; G g; H h; P p; T d1; E d0;
  int n;
  vector<int> dat, lazy;

  segtree(){}
  segtree(int n_, F f_, G g_, H h_, T d1_, E d0_, P p_=[](E a, int b){return a;}):
    f(f_), g(g_), h(h_), p(p_), d1(d1_), d0(d0_) {
    n = 1; while(n < n_) n *= 2;
    dat.assign(n*2, d1);
    lazy.assign(n*2, d0);
  }
  void build(vector<T> v) {
    REP(i, v.size()) dat[i+n-1] = v[i];
    for(int i=n-2; i>=0; --i) dat[i] = f(dat[i*2+1], dat[i*2+2]);
  }

  // 区間の幅がlenの節点kについて遅延評価
  inline void eval(int len, int k) {
    if(lazy[k] == d0) return;
    if(k*2+1 < n*2-1) {
      lazy[2*k+1] = h(lazy[k*2+1], lazy[k]);
      lazy[2*k+2] = h(lazy[k*2+2], lazy[k]);
    }
    dat[k] = g(dat[k],p(lazy[k],len));
    lazy[k] = d0;
  }
  // [a, b)
  T update(int a, int b, ll x, int k, int l, int r) {
    eval(r-l, k);
    if(b <= l || r <= a) return dat[k];
    if(a <= l && r <= b) {
      lazy[k] = h(lazy[k], x);
      return g(dat[k], p(lazy[k],r-l));
    }
    return dat[k] = f(update(a, b, x, 2*k+1, l, (l+r)/2),
                      update(a, b, x, 2*k+2, (l+r)/2, r));
  }
  void update(int a, int b, ll x) { update(a, b, x, 0, 0, n); }
  // [a, b)
  ll query(int a, int b, int k, int l, int r) {
    eval(r-l, k);
    if(b <= l || r <= a) return INT_MAX;
    if(a <= l && r <= b) return dat[k];
    T vl = query(a, b, 2*k+1, l, (l+r)/2);
    T vr = query(a, b, 2*k+2, (l+r)/2, r);
    return f(vl, vr);
  }
  ll query(int a, int b) { return query(a, b, 0, 0, n); }
  // デバッグ出力
  void debug() {
    cout << "---------------------" << endl;
    int cnt = 0;
    for(int i=1; i<=n; i*=2) {
      REP(j, i) {cout << "(" << dat[cnt] << "," << lazy[cnt] << ") "; cnt++;}
      cout << endl;
    }
    cout << "---------------------" << endl;
  }
};

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n, q;
  cin >> n >> q;

  auto f = [](int l, int r){return min(l,r);};
  auto g = [](int l, int r){return l+r;};
  auto h = [](int l, int r){return l+r;};
  segtree<int,int> seg(n, f, g, h, 0, 0);

  REP(i, q) {
    int type; cin >> type;
    if(type == 0) {
      int s, t, x;
      cin >> s >> t >> x;
      seg.update(s, t+1, x);
    } else {
      int s, t;
      cin >> s >> t;
      cout << seg.query(s, t+1) << endl;
    }
    // seg.debug();
  }

  return 0;
}

