同一个S可能需要多次经过，只需杀一次。

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;const int N = 100 + 5;char s[N][N];
int si, sj, n, m, ei, ej, p[N][N];struct Node {int x, y, key, dis, snack;Node(int x = 0, int y = 0, int key = 0, int dis = 0, int snack = 0) :x(x), y(y), key(key), dis(dis), snack(snack) {}bool operator<(const Node& n)const {return dis > n.dis;}
};int dir[][4] = { {-1,1,0,0},{0,0,-1,1} };
int d[N][N][10];int bfs() {priority_queue<Node> q;for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)for (int k = 0; k <= m; k++)d[i][j][k] = 0x3f3f3f3f;q.push(Node(si, sj, 0, 0, 0));d[si][sj][0] = 0;while (!q.empty()) {Node tp = q.top();q.pop();for (int i = 0; i < 4; i++) {int ii = tp.x + dir[0][i];int jj = tp.y + dir[1][i];if (0 <= ii && ii < n && 0 <= jj && jj < n && s[ii][jj] != '#') {if (s[ii][jj] == 'S') {if ((tp.snack >> p[ii][jj]) & 1){if (tp.dis + 1 < d[ii][jj][tp.key])d[ii][jj][tp.key] = tp.dis + 1, q.push(Node(ii, jj, tp.key, tp.dis + 1, tp.snack));}else {if (tp.dis + 2 < d[ii][jj][tp.key])d[ii][jj][tp.key] = tp.dis + 2, q.push(Node(ii, jj, tp.key, tp.dis + 2, tp.snack | (1 << p[ii][jj])));}}else if ('1' <= s[ii][jj] && s[ii][jj] <= '9') {if (s[ii][jj] - '0' == tp.key + 1) {if (tp.dis + 1 < d[ii][jj][tp.key + 1]) {d[ii][jj][tp.key + 1] = tp.dis + 1, q.push(Node(ii, jj, tp.key + 1, tp.dis + 1, tp.snack));}}else {if (tp.dis + 1 < d[ii][jj][tp.key])d[ii][jj][tp.key] = tp.dis + 1, q.push(Node(ii, jj, tp.key, tp.dis + 1, tp.snack));}}else if (s[ii][jj] == 'T') {if (tp.key == m)return tp.dis + 1;else{if (tp.dis + 1 < d[ii][jj][tp.key])d[ii][jj][tp.key] = tp.dis + 1, q.push(Node(ii, jj, tp.key, tp.dis + 1, tp.snack));}}else {if (tp.dis + 1 < d[ii][jj][tp.key])d[ii][jj][tp.key] = tp.dis + 1, q.push(Node(ii, jj, tp.key, tp.dis + 1, tp.snack));}}}}return -1;
}int main(){for (; scanf("%d%d", &n, &m) == 2 && n + m;) {si = -1, sj = -1, ei = -1, ej = -1;for (int i = 0; i < n; i++)scanf("%s", s[i]);int cnt = 0;for(int i=0;i<n;i++)for (int j = 0; j < n; j++){if (s[i][j] == 'K')si = i, sj = j;else if (s[i][j] == 'T')ei = i, ej = j;else if (s[i][j] == 'S') p[i][j] = cnt++;}if (ei == -1 || si == -1) {printf("impossible\n");}else {int ans = bfs();if (ans != -1) printf("%d\n", ans);else printf("impossible\n");}}return 0;
}