Problem:
Quick Sort

Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode:

Partition(A, p, r)
1 x = A[r]
2 i = p-1
3 for j = p to r-1
4 do if A[j] <= x
5 then i = i+1
6 exchange A[i] and A[j]
7 exchange A[i+1] and A[r]
8 return i+1

Quicksort(A, p, r)
1 if p < r
2 then q = Partition(A, p, r)
3 run Quicksort(A, p, q-1)
4 run Quicksort(A, q+1, r)

Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers.

Your program should also report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).

Input

The first line contains an integer n, the number of cards.

n cards are given in the following lines. Each card is given in a line and represented by a pair of a character and an integer separated by a single space.

Output

In the first line, print the stability ("Stable" or "Not stable") of this output.

In the following lines, print the arranged cards in the same manner of that of the input.

Constraints

1 ≤ n ≤ 100,000

1 ≤ the number of a card ≤ 109

There are no identical card in the input

Sample Input 1

6
D 3
H 2
D 1
S 3
D 2
C 1

Sample Output 1

Not stable
D 1
C 1
D 2
H 2
D 3
S 3

Sample Input 2

2
S 1
H 1

Sample Output 2

Stable
S 1
H 1

Solution in C (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define SENTINEL 2000000000

long int cnt=0;

typedef struct box{
	char a;
	int b;
} card;

void merge(card *A,int left,int mid,int right)//分割した配列を小さい順に結合
{
  int n1 = mid - left;
  int n2 = right - mid;
  card L[n1+1],R[n2+1];
  int k,i,j;
  for (i=0;i<n1;i++) {
  	L[i].a = A[left + i].a;
  	L[i].b = A[left + i].b;
  }
  for (i=0;i<n2;i++) {
  	R[i].a = A[mid + i].a;
  	R[i].b = A[mid + i].b;
  }
  L[n1].b = SENTINEL;//EOF
  R[n2].b = SENTINEL;//EOF
  i = 0;
  j = 0;
    
  for(k=left;k<right;k++){//iが左の要素番号,jが右の要素番号
    if(L[i].b<=R[j].b){//左が小さい場合
      A[k].a = L[i].a;
      A[k].b = L[i].b;
      i=i+1;
    }else{ //右が小さい場合
      A[k].a = R[j].a;
       A[k].b = R[j].b;
      if(L[i].b!=SENTINEL) cnt+=(n1-i);//SENTINELと比較する場合を除いて左側の未ソート部分の要素数をcntに加算
      j = j + 1;
    }
  }
}

void mergeSort(card *A,int left,int right)
{
  int mid,i;
  if(left+1<right){//二分探索のように左端が右端を超えたら終了
    mid = (left + right)/2;
    mergeSort(A, left, mid);//左半分で再帰
    mergeSort(A, mid, right);//右半分で再帰
    merge(A, left, mid, right);//結合
  }
}


int partition(card *A,int p,int r)
{
	int x = A[r].b;//基準値をxに格納
	int box,j,i = p-1;
	
	int k;
	
	char s;
	for( j = p ; j < r ; j++){//jがこれから探索する要素番号を示す
		if( A[j].b <= x){//xより小さければ
			i = i+1;//iをインクリメントする
			box=A[i].b;//xより大きくかつ最後に追加した要素と入れ替える
			s=A[i].a;
			A[i].b=A[j].b;
			A[i].a=A[j].a;
			A[j].b=box;
			A[j].a=s;
		}
	}
	box=A[i+1].b;//基準値をiの位置に格納する
	s=A[i+1].a;
	A[i+1].b=A[r].b;
	A[i+1].a=A[r].a;
	A[r].b=box;
	A[r].a=s;
	//for(k=p;k<r;k++) printf("%c %d\n",A[k].a,A[k].b);
	return i+1;
}


void quicksort(card *A,int p ,int r)
{
	int q;
	if(p<r){
		q=partition(A,p,r);
		quicksort(A,p,q-1);
		quicksort(A,q+1,r);
	}
}

int main(void)
{
	int i,n;
	card *A;
	card *B;
	int left,right;
	int check=0;
	scanf("%d",&n);
	A = (card*)malloc(sizeof(card)*n+1);
	B = (card*)malloc(sizeof(card)*n+1);
	for(i=0;i<n;i++) {
		scanf("%s",&A[i].a);
		scanf("%d",&A[i].b);
		B[i].a=A[i].a;
		B[i].b=A[i].b;
	}
	
	quicksort(A,0,n-1);
	 left=0;
    right=n;
	mergeSort(B,left,right);
	for(i=0;i<n;i++) if(A[i].a!=B[i].a) check=1;
	if(check==0)printf("Stable\n");
	else printf("Not stable\n");
	for(i=0;i<n;i++) printf("%c %d\n",A[i].a,A[i].b);
	free(A);
	free(B);
	return 0;
}