#include <stdio.h>
 
int main() {
	int k, i, j, temp, instancia = 0, solucao = 1;
	int matriz[9][9], somaLinha[9], somaColuna[9], somaMatriz[3][3];
 
	scanf("%i", &k);
 
    for(int cont = 0; cont < k; cont++) {
    	//leitura de valores para a matriz
		for(i = 0; i < 9; i++) {
			for(j = 0; j < 9; j++) { 
				scanf("%i", &matriz[i][j]);
			}
    	}
 
		//inicializa variáveis somadoras com zero
		for (i = 0; i < 9; i++) {
			somaLinha[i]         = 0;
			somaColuna[i]        = 0;
			somaMatriz[i/3][i%3] = 0;
			//descomente o prinf abaixo para facilitar o entendimento sobre
			//como a matriz está sendo preenchida sem utilizar dois for 
			//printf("%i %i \n", i/3, i%3);
		}
 
    	//contabiliza valores para resolver o problema
		for (i = 0; i < 9; i++) {
			for (j = 0; j < 9; j++) {
				temp = matriz[i][j] * matriz[i][j];
				somaLinha[i]  += temp;
				somaColuna[j] += temp;
				somaMatriz[i/3][j/3] += temp;
				//printf("%i %i %i %i %i\n", i, j, i/3, j/3, matriz[i][j]);
 
				if (i == 8 && somaColuna[j] != 285) { break; }
			} 
 
			if (somaLinha[i] != 285) { break; }
		}
 
		//mostra os resultados
		instancia++;
		printf("Instancia %i\n", instancia);
 
		//determina se é a solução
		for (int i = 0; i < 3; i++) {
			for (int j = 0; j < 3; j++) {
				if (somaMatriz[i][j] != 285) {
					solucao = 0;
				}
			}
    	}
 
		if (solucao == 1) {
			printf("SIM\n\n");
		} else {
			printf("NAO\n\n");
		}
 
		solucao = 1;
	}
 
    return 0;
}

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