Problem: E: Cut out the sum

problem

Ebi-chan has a sequence (a_1, a_2, ..., a_n). Ebi-chan is worried about the maximum value of the sum of the partial arrays (which may be empty), so please find it. Here, a subarray refers to a continuous subsequence. The sum of empty strings is 0.

In addition, Ebi-chan rewrites this sequence q times, so find this value immediately after each rewrite.

Input format


n q
a_1 a_2 ... a_n
k_1 x_1
...
k_q x_q


Each k_j, x_j (1 \ leq j \ leq q) means rewrite a_ {k_j} to x_j. Note that each rewrite is not independent and the array remains rewritten in subsequent processing.

Constraint

* 1 \ leq n \ leq 10 ^ 5
* 1 \ leq q \ leq 10 ^ 5
* 1 \ leq k_j \ leq n
* | a_i |, | x_j | \ leq 10 ^ 9



Output format

Please output q + 1 line. Output the maximum value of the sum of the subarrays before rewriting on the first line and immediately after the i-th rewrite on the 1 + i line.

Input example 1


5 2
1 2 -3 4 -5
3 3
2 -6


Output example 1


Four
Ten
7


If the subarray (a_l,…, a_r) is represented as a [l, r], the sum of a [1, 4] and a [4, 4] is 4 before rewriting, which is the maximum.

After the first rewrite, the sequence becomes (1, 2, 3, 4, -5), and the maximum sum is 10 of a [1, 4].

After the second rewrite, the sequence becomes (1, -6, 3, 4, -5) and the maximum sum is 7 in a [3, 4].

Input example 2


3 1
-one two Three
1 1


Output example 2


0
1


The sum of the empty strings is 0, which is the maximum before rewriting.





Example

Input

5 2
1 2 -3 4 -5
3 3
2 -6


Output

4
10
7

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef long double D;
typedef complex<D> P;
#define F first
#define S second
const ll MOD=1000000007;
//const ll MOD=998244353;

template<typename T,typename U>istream & operator >> (istream &i,pair<T,U> &A){i>>A.F>>A.S; return i;}
template<typename T>istream & operator >> (istream &i,vector<T> &A){for(auto &I:A){i>>I;} return i;}
template<typename T,typename U>ostream & operator << (ostream &o,const pair<T,U> &A){o<<A.F<<" "<<A.S; return o;}
template<typename T>ostream & operator << (ostream &o,const vector<T> &A){ll i=A.size(); for(auto &I:A){o<<I<<(--i?" ":"");} return o;}


//BEGIN CUT HERE
template <typename T,typename E>
struct SegmentTree{
  using F = function<T(T,T)>;
  using G = function<T(T,E)>;
  using H = function<E(E,E)>;
  int n,height;
  F f;
  G g;
  H h;
  T ti;
  E ei;
  vector<T> dat;
  vector<E> laz;
  SegmentTree(F f,G g,H h,T ti,E ei):
    f(f),g(g),h(h),ti(ti),ei(ei){}

  void init(int n_){
    n=1;height=0;
    while(n<n_) n<<=1,height++;
    dat.assign(2*n,ti);
    laz.assign(2*n,ei);
  }
  void build(const vector<T> &v){
    int n_=v.size();
    init(n_);
    for(int i=0;i<n_;i++) dat[n+i]=v[i];
    for(int i=n-1;i;i--)
      dat[i]=f(dat[(i<<1)|0],dat[(i<<1)|1]);
  }
  inline T reflect(int k){
    return laz[k]==ei?dat[k]:g(dat[k],laz[k]);
  }
  inline void eval(int k){
    if(laz[k]==ei) return;
    laz[(k<<1)|0]=h(laz[(k<<1)|0],laz[k]);
    laz[(k<<1)|1]=h(laz[(k<<1)|1],laz[k]);
    dat[k]=reflect(k);
    laz[k]=ei;
  }
  inline void thrust(int k){
    for(int i=height;i;i--) eval(k>>i);
  }
  inline void recalc(int k){
    while(k>>=1)
      dat[k]=f(reflect((k<<1)|0),reflect((k<<1)|1));
  }
  void update(int a,int b,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);
  }
  void set_val(int a,T x){
    thrust(a+=n);
    dat[a]=x;laz[a]=ei;
    recalc(a);
  }
  T query(int a,int b){
    thrust(a+=n);
    thrust(b+=n-1);
    T vl=ti,vr=ti;
    for(int l=a,r=b+1;l<r;l>>=1,r>>=1) {
      if(l&1) vl=f(vl,reflect(l++));
      if(r&1) vr=f(reflect(--r),vr);
    }
    return f(vl,vr);
  }

  template<typename C>
  int find(int st,C &check,T &acc,int k,int l,int r){
    if(l+1==r){
      acc=f(acc,reflect(k));
      return check(acc)?k-n:-1;
    }
    eval(k);
    int m=(l+r)>>1;
    if(m<=st) return find(st,check,acc,(k<<1)|1,m,r);
    if(st<=l&&!check(f(acc,dat[k]))){
      acc=f(acc,dat[k]);
      return -1;
    }
    int vl=find(st,check,acc,(k<<1)|0,l,m);
    if(~vl) return vl;
    return find(st,check,acc,(k<<1)|1,m,r);
  }
  template<typename C>
  int find(int st,C &check){
    T acc=ti;
    return find(st,check,acc,1,0,n);
  }
};
//END CUT HERE

struct node{
  ll mi,mx,ans,sum;
  node(ll a=0,ll b=0,ll c=0,ll d=0):mi(a),mx(b),ans(c),sum(d){}

  bool operator == (const node &A){return A.mi==mi && A.mx==mx && A.ans==ans && A.sum==sum;}
};

const ll uku=-1*MOD;

node err=node(MOD,uku,0,0);

int main(){
  auto f=[](node a,node b){
           return node(min(a.mi,a.sum+b.mi),max(a.mx,a.sum+b.mx),max(max(a.ans,b.ans),a.sum+b.mx-a.mi),a.sum+b.sum);
         };
  auto g=[](node a,node b){return b==err?a:b;};
  auto h=[](node a,node b){return b==err?a:b;};

  SegmentTree<node,node> ch(f,g,h,err,err);

  ll N,Q;
  cin>>N>>Q;
  N++;
  ch.build(vector<node>(N,node()));
  vector<ll> A(N-1);
  cin>>A;
  for(int i=1;i<=N;i++){ch.update(i,i+1,node(A[i-1],A[i-1],0,A[i-1]));}
  cout<<ch.query(0,N).ans<<endl;
  while(Q--){
    /*for(int i=0;i<N;i++){
      node ret=ch.query(i,i+1);
      cout<<ret.mi<<" "<<ret.mx<<" "<<ret.ans<<" "<<ret.sum<<endl;
      }*/
    ll k,x;
    cin>>k>>x;
    ch.update(k,k+1,node(x,x,0,x));
    cout<<ch.query(0,N).ans<<endl;
  }

  return 0;
}