/*思路:把输入的每一组电话号码改写成interger,由函数trsnb(转换成数字)
负责.然后把这一组数字中的最小者取出来.然后一个个对照这排数.如果相同的
话,把相同的数改成-1,同时修改相应的判断参数.这样判一组相同的数打一个.最后如果
都没打印过的话,就输出一个No duplicates.*/
#include <stdio.h>
#include <malloc.h>
int po(int a, int b){
	int x,result=1;
	for(x=0; x<b; x++){
		result*=a;
	}
	return result;
}
int trsnb(char s[16], int od){
	if (s[od]>='A' && s[od]<'Z'){
		if (s[od]>='Q')
			s[od]--;
		return (s[od]-59)/3;
	}
	else if(s[od]>='0' && s[od]<='9'){
		return (s[od]-48);
	}
}
struct k{
	char a[16];
}*ip;//input dials and (*)is often neglected

void main(){
	int *st;//it is used to store transformed dials.
	int n,i,ca,m,j,min,num;
	scanf("%d",&ca);
	ip=(struct k*)malloc(sizeof(struct k)*(ca+1));//apply for room
	st=(int *)malloc(sizeof(int)*(ca+1));//warning st,ip begin with 1
	for (n=1; n<=ca; n++){
		scanf("%s",&(*(ip+n)).a);
		*(st+n)=0;
		j=1;//j for 1-7
		for (i=0; (*(ip+n)).a[i]!='\0'; i++){
			if ((*(ip+n)).a[i]!='-'){
				*(st+n)+=trsnb((*(ip+n)).a,i)*po(10,(7-j));
				j++;
			}
		}
	}//all dials have been transformed to intergers.
	int q=ca;
	for (m=ca; m!=0;){
		min=9999999;
		num=0;//menbers of a group
		for (n=1; n<=ca; n++){
			if(*(st+n)!=0){
				if(*(st+n)<min)
					min=*(st+n);
			}
		}//find the min one of the group
		for (n=1; n<=ca; n++){
			if(*(st+n)==min){
				*(st+n)=0;
				num++;
				m--;
			}
		}
		if(num>1){
			printf("%d-%d %d\n",min/10000,min%10000,num);
			q--;
		}
	}
	if(q==ca) printf("No duplicates.\n");
}