(10%) The following procedure is a bottom-up method for matrix chain order. This procedure assumes that matrix $A_i$ has dimensions $p_{i-1} \times p_i$ for $i = 1, 2, ..., n$. Its input is a sequence $p = \langle p_0, p_1, ..., p_n \rangle$, where $p.length = n+1$. Let $m[i,j]$ be the minimum number of scalar multiplications needed to compute the matrix $A_i A_{i+1} \cdots A_j$. The procedure uses an auxiliary table $m[1..n, 1..n]$ for storing the $m[i,j]$ costs and another auxiliary table $s[1..n-1, 2..n]$ that records which index of $k$ achieved the optimal cost in computing $m[i,j]$. Please fill in the empty statements.

n = p.length - 1
let m[1..n,1..n] and s[1..n-1,2..n] be new tables
for i = 1 to n
    m[i,i] = 0
for l = 2 to n   // l is the chain length
    for i = 1 to n-l+1
        j = ____(a) (3%)____
        m[i,j] = infinity
        for k = i to j-1
            q = ____(b) (4%)____
            if q < m[i,j]
                m[i,j] = q
                s[i,j] = ____(c) (3%)____