Problem: Range Min of Max Query

Given a sequence of integer pairs (a_1, b_1), (a_2, b_2), .., (a_N, b_N).

Handle two types of queries.

The first type of query adds X to a_L, a_ {L + 1}, .., a_R.

The second type of query finds the minimum values ​​of max (a_L, b_L), max (a_ {L + 1}, b_ {L + 1}), .., max (a_R, b_R).

input


N Q
a_1 b_1
a_2 b_2
::
a_N b_N
query_1_1
query_2
::
query_Q


If the i-th query is the first type of query, query_i will be 1 L_i R_i X_i.

If the i-th query is the second type of query, query_i will be 2 L_i R_i.

output


ans_1
ans_2
::
ans_k


Output the answers to the second type of query in order.

Constraint

* 1 \ leq N, Q \ leq 10 ^ 5
* 1 \ leq a_i, b_i \ leq 10 ^ 9
* 1 \ leq L_i \ leq R_i \ leq N
* -10 ^ 9 \ leq X_i \ leq 10 ^ 9



Input example


6 6
8 1
6 1
9 4
1 5
twenty one
14
2 1 3
1 1 3 3
2 1 3
2 4 6
1 4 6 3
2 4 6


Output example


6
9
2
Four






Example

Input

6 6
8 1
6 1
9 4
1 5
2 1
1 4
2 1 3
1 1 3 3
2 1 3
2 4 6
1 4 6 3
2 4 6


Output

6
9
2
4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define rep(i,N) for(long long i = 0; i < (long long)(N); i++)
#define repr(i,N) for(long long i = (long long)(N) - 1; i >= 0; i--)
#define rep1(i,N) for(long long i = 1; i <= (long long)(N) ; i++)
#define repr1(i,N) for(long long i = (N) ; (long long)(i) > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
using namespace std; void solve();
using ll = long long; template<class T = ll> using V = vector<T>;
using vi = V<int>; using vl = V<>; using vvi = V< V<int> >;
constexpr int inf = 1001001001; constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,N) do { cerr << #v << " = "; array_out(v , N);cout << endl;} while(0)
#else
  #define trc(...)
  #define trca(...)
  int main(){solve();}
#endif

using P = pair<ll,ll>; using vp = V<P>;
constexpr int MOD = /**/ 1000000007; //*/ 998244353;
////////////////////////////////

constexpr int B = 300;

struct block{
  ll x;
  vl a,b,amb,ra,rb;
  block() 
    : x(0) , a(B , 0) , b(B , infLL) , amb(B) , ra(B + 1) , rb(B + 1){}

  void build(){
    vi ord(B); iota(all(ord) , 0);
    sort(all(ord),[&](int x,int y){
      return a[x] - b[x] < a[y] - b[y];
    });
    rep(i , B) amb[i] = a[ord[i]] - b[ord[i]];
    ra[B] = infLL;
    repr(i , B) ra[i] = min(ra[i + 1] , a[ ord[i] ]);
    rb[0] = infLL;
    rep1(i , B) rb[i] = min(rb[i - 1] , b[ ord[i - 1] ]);
  }

  void update_all(ll _x){
    x += _x;
  }
  void update_part(int l,int r,ll x){
    for(int i = l ; i < r ; i++) a[i] += x;
    build();
  }

  ll query_all(){
    auto it = lower_bound(all(amb) , -x);
    int idx = it - amb.begin();
    // it以降 a - b >= -x <-> a + x >= b
    // it未満 a - b <  -x <-> a + x <  b
    // よって aのidx以上とbのidx未満の最小を見ればよい
    return min(rb[idx] , ra[idx] + x); 
  }
  ll query_part(int l,int r){
    ll ret = infLL;
    for(int i = l ; i < r ; i++){
      amin(ret , max(a[i] + x , b[i]));
    }
    return ret;
  }
};

void solve(){
  ini(N , Q);
  V<block> sq(N / B + 1);
  rep(i , N) in(sq[i / B].a[i % B] , sq[i / B].b[i % B]);
  each(x , sq) x.build();

  auto query = [&](int l,int r) -> ll{
    if(l / B == r / B){
      return sq[l / B].query_part(l % B , r % B);
    }
    else {
      ll ret = infLL;
      amin(ret , sq[l / B].query_part(l % B , B) );
      for(int i = l / B + 1 ; i < r / B ; i++) amin(ret , sq[i].query_all() );
      amin(ret , sq[r / B].query_part(0 , r % B ) );
      return ret;
    }
  };

  auto update = [&](int l,int r , ll x) -> void{
    if(l / B == r / B){
      sq[l / B].update_part(l % B , r % B , x);
    }
    else {
      sq[l / B].update_part(l % B , B , x);
      for(int i = l / B + 1 ; i < r / B ; i++) sq[i].update_all(x);
      sq[r / B].update_part(0 , r % B , x);
    }
  };

  rep(i , Q){
    ini(c);
    if(c == 1){
      inl(l , r , x);
      update(l - 1, r , x);
    }
    else {
      ini(l , r);
      out(query(l - 1 , r));
    }
  }



}