Phoenix is trying to take a photo of his n friends with labels 1, 2, ..., n who are lined up in a row in a special order. But before he can take the photo, his friends get distracted by a duck and mess up their order.

Now, Phoenix must restore the order but he doesn't remember completely! He only remembers that the i-th friend from the left had a label between a_i and b_i inclusive. Does there exist a unique way to order his friends based of his memory? 

Input

The first line contains one integer n (1 ≤ n ≤ 2⋅10^5) — the number of friends.

The i-th of the next n lines contain two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ n) — Phoenix's memory of the i-th position from the left.

It is guaranteed that Phoenix's memory is valid so there is at least one valid ordering.

Output

If Phoenix can reorder his friends in a unique order, print YES followed by n integers — the i-th integer should be the label of the i-th friend from the left.

Otherwise, print NO. Then, print any two distinct valid orderings on the following two lines. If are multiple solutions, print any.

Examples

Input


4
4 4
1 3
2 4
3 4


Output


YES
4 1 2 3 


Input


4
1 3
2 4
3 4
2 3


Output


NO
1 3 4 2 
1 2 4 3 
4
3 4
1 2
3 4
1 2
 4
2 4
1 3
4 4
1 1
 10
6 6
1 2
2 2
6 7
8 8
3 5
5 5
4 4
8 9
10 10
 4
1 4
1 4
2 4
1 2
 8
3 4
7 8
2 6
1 3
6 7
5 8
4 6
8 8
 7
7 7
2 3
2 3
5 7
5 6
4 4
1 2
 15
2 2
10 11
2 5
6 7
7 12
9 9
2 3
5 14
1 3
10 10
15 15
12 12
5 6
6 6
14 15
 5
1 3
4 4
5 5
1 4
1 5
 4
1 4
2 3
2 3
4 4
 4
1 4
1 4
1 2
1 4
 5
1 3
1 3
1 5
3 4
1 5
 3
1 3
2 2
1 3
 5
4 4
1 4
3 4
1 3
3 5
 3
1 3
1 2
2 3
 4
2 4
1 3
2 3
1 2
 5
4 5
4 5
1 3
1 2
2 4
 6
2 6
4 6
6 6
2 2
3 3
1 3
 3
2 3
1 2
1 3
 6
4 6
1 3
4 6
4 4
1 4
1 2
 7
3 6
6 6
5 7
2 4
2 3
1 2
3 5
 8
4 6
8 8
1 4
4 5
6 8
1 2
3 6
1 4
 6
2 4
2 6
4 5
2 5
6 6
1 4
 8
4 4
7 7
3 5
5 7
1 8
4 5
1 5
2 2
 4
3 4
1 3
2 3
1 2
 10
9 9
3 4
1 1
5 5
7 7
3 7
6 10
3 3
10 10
1 3
 4
1 4
4 4
2 3
1 1
 1
1 1
 8
1 5
6 8
1 3
4 8
1 1
3 3
6 6
8 8
 5
3 3
4 4
2 2
1 1
5 5
 4
1 1
2 3
2 3
1 4
 4
3 4
1 4
2 3
1 2
 2
1 2
1 2
 8
3 3
3 5
2 3
1 1
5 7
7 7
5 5
3 8
 5
2 2
3 3
1 2
1 5
4 4
 6
1 3
2 4
3 4
2 3
5 5
6 6
 6
1 4
3 6
6 6
3 5
1 5
2 5
 5
4 4
2 3
4 5
1 4
2 4
 3
1 2
1 2
1 3
 4
1 4
2 2
3 3
1 4
 5
2 5
1 1
2 3
3 4
5 5
 4
2 4
1 4
3 4
1 3
 6
1 1
4 6
3 6
1 3
3 4
3 6
 5
2 3
3 4
4 5
1 1
2 3
 5
2 3
1 5
2 4
3 5
4 4
 20
12 14
13 13
18 18
3 3
7 8
11 11
10 13
2 2
9 10
3 5
7 8
15 17
14 15
9 9
20 20
15 16
6 6
19 19
1 1
5 5
 5
1 5
2 2
3 3
4 4
1 5
 7
5 7
1 5
1 2
1 5
5 6
3 5
7 7

NO
3 1 4 2
3 2 4 1
 NO
3 2 4 1 
2 3 4 1  YES
6 1 2 7 8 3 5 4 9 10  NO
2 3 4 1 
1 3 4 2 
 YES
3 7 2 1 6 5 4 8  NO
7 2 3 6 5 4 1 
7 3 2 6 5 4 1 
 YES
2 11 4 7 8 9 3 13 1 10 15 12 5 6 14  NO
1 4 5 2 3 
2 4 5 1 3  NO
1 2 3 4 
1 3 2 4 
 NO
2 3 1 4 
1 3 2 4 
 NO
1 2 4 3 5 
2 1 4 3 5 
 NO
1 2 3 
3 2 1 
 NO
4 2 3 1 5 
4 1 3 2 5  NO
2 1 3
1 2 3
 NO
4 2 3 1
4 1 3 2
 NO
4 5 2 1 3
4 5 1 2 3
 NO
4 5 6 2 3 1 
5 4 6 2 3 1 
 NO
2 1 3 
3 1 2 
 NO
5 2 6 4 3 1 
5 1 6 4 3 2 
 NO
5 6 7 3 2 1 4 
5 6 7 2 3 1 4  NO
5 8 2 4 7 1 6 3
5 8 1 4 7 2 6 3
 NO
2 5 4 3 6 1 
3 5 4 2 6 1  YES
4 7 3 6 8 5 1 2  NO
4 2 3 1
4 1 3 2
 YES
9 4 1 5 7 6 8 3 10 2  NO
3 4 2 1 
2 4 3 1  YES
1  NO
4 7 2 5 1 3 6 8 
5 7 2 4 1 3 6 8  YES
3 4 2 1 5  NO
1 2 3 4 
1 3 2 4 
 NO
3 4 2 1 
4 3 2 1 
 NO
1 2 
2 1 
 YES
3 4 2 1 6 7 5 8  YES
2 3 1 5 4  NO
1 3 4 2 5 6
1 2 4 3 5 6
 NO
1 5 6 3 2 4 
2 5 6 3 1 4 
 NO
4 2 5 1 3 
4 3 5 1 2  NO
1 2 3 
2 1 3 
 NO
1 2 3 4 
4 2 3 1 
 NO
4 1 2 3 5 
3 1 2 4 5 
 NO
2 3 4 1
2 1 4 3
 NO
1 4 5 2 3 6 
1 5 4 2 3 6 
 NO
2 4 5 1 3 
3 4 5 1 2 
 NO
2 1 3 5 4 
3 1 2 5 4  NO
14 13 18 3 7 11 12 2 10 4 8 17 15 9 20 16 6 19 1 5 
14 13 18 3 8 11 12 2 10 4 7 17 15 9 20 16 6 19 1 5 
 NO
1 2 3 4 5 
5 2 3 4 1 
 NO
6 2 1 3 5 4 7
6 1 2 3 5 4 7

//#define _GLIBCXX_DEBUG

#include<bits/stdc++.h>
using namespace std;

#define endl '\n'
#define lfs cout<<fixed<<setprecision(10)
#define ALL(a)  (a).begin(),(a).end()
#define ALLR(a)  (a).rbegin(),(a).rend()
#define spa << " " <<
#define fi first
#define se second
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define EB emplace_back
#define rep(i,n,m) for(ll i = (n); i < (ll)(m); i++)
#define rrep(i,n,m) for(ll i = (ll)(m) - 1; i >= (ll)(n); i--)
using ll = long long;
using ld = long double;
const ll MOD1 = 1e9+7;
const ll MOD9 = 998244353;
const ll INF = 1e18;
using P = pair<ll, ll>;
template<typename T1, typename T2>
bool chmin(T1 &a,T2 b){if(a>b){a=b;return true;}else return false;}
template<typename T1, typename T2>
bool chmax(T1 &a,T2 b){if(a<b){a=b;return true;}else return false;}
ll median(ll a,ll b, ll c){return a+b+c-max({a,b,c})-min({a,b,c});}
void ans1(bool x){if(x) cout<<"Yes"<<endl;else cout<<"No"<<endl;}
void ans2(bool x){if(x) cout<<"YES"<<endl;else cout<<"NO"<<endl;}
void ans3(bool x){if(x) cout<<"Yay!"<<endl;else cout<<":("<<endl;}
template<typename T1,typename T2>
void ans(bool x,T1 y,T2 z){if(x)cout<<y<<endl;else cout<<z<<endl;}  
template<typename T>
void debug(vector<vector<T>>&v,ll h,ll w){for(ll i=0;i<h;i++)
{cout<<v[i][0];for(ll j=1;j<w;j++)cout spa v[i][j];cout<<endl;}};
void debug(vector<string>&v,ll h,ll w){for(ll i=0;i<h;i++)
{for(ll j=0;j<w;j++)cout<<v[i][j];cout<<endl;}};
template<typename T>
void debug(vector<T>&v,ll n){if(n!=0)cout<<v[0];
for(ll i=1;i<n;i++)cout spa v[i];cout<<endl;};
template<typename T>
vector<vector<T>>vec(ll x, ll y, T w){
  vector<vector<T>>v(x,vector<T>(y,w));return v;}
ll gcd(ll x,ll y){ll r;while(y!=0&&(r=x%y)!=0){x=y;y=r;}return y==0?x:y;}
vector<ll>dx={1,0,-1,0,1,1,-1,-1};
vector<ll>dy={0,1,0,-1,1,-1,1,-1};
template<typename T>
vector<T> make_v(size_t a,T b){return vector<T>(a,b);}
template<typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v(ts...))>(a,make_v(ts...));
}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2>&p){
  return os << p.first << " " << p.second;
}  
int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  ll res=0,buf=0;
  bool judge = true;
  ll n,k;cin>>n>>k;
  auto dp=vec(n+1,k,false);
  dp[0][0]=true;
  vector<ll>a(n),b(n);
  rep(i,0,n)cin>>a[i]>>b[i];
  ll suma=0,sumb=0;
  rep(i,0,n){
    vector<bool>t(k);
    rep(j,0,k){
      if(a[i]<j||((a[i]-j)%k!=0&&(a[i]-j)%k+b[i]<k))continue;
      t[j]=true;
    }
    //debug(t,k);
    //cout<<mi spa mx<<endl;
    rep(j,0,k){
      if(!dp[i][j])continue;
      ll tmp=j;
      rep(o,0,k){
        if(!t[o])continue;
        dp[i+1][tmp]=true;
      }
    }
    suma+=a[i],sumb+=b[i];
  }
  res=suma/k+sumb/k;
  //debug(dp,n+1,k);
  rep(i,0,k){
    if(!dp[n][i]||suma%k+sumb%k<k)continue;
    ll db=(sumb-(k-(suma-i)%k)%k)%k;
    //cout<<i spa db<<endl;
    chmax(res,suma/k+sumb/k+(db+i<k));
  }
  cout<<res<<endl;
  return 0;
}