In the figure, a piece of paper PQRS in the shape of a parallelogram is folded along 2 dotted lines, TS and UR as shown in Figure 1. In Figure 2, $\angle PST = 32^\circ$ and $\angle QRS = 14^\circ$.

1. Relate the folded angle $\angle UQR$ to the original angle of the parallelogram.

When the paper is folded, the angle $\angle UQR$ in Figure 2 corresponds directly to the original interior angle $\angle PQR$ of the parallelogram PQRS before folding because the shape around Q is preserved under folding.

Since PQRS is a parallelogram, the opposite interior angles are equal:
$\angle PQR = \angle PSR$

2. Calculate the original interior angle $\angle PSR$ to find $\angle UQR$.

The fold line $TS$ is perpendicular to the parallel horizontal sides $PQ$ and $SR$ of the parallelogram. Therefore, the angle between the vertical line $TS$ and the base $SR$ is $90^\circ$ (i.e., $\angle TSR = 90^\circ$).

The full angle at $S$ before folding, $\angle PSR$, is the sum of the angle of the folded part and the right angle:
$\angle PSR = \angle PST + \angle TSR = 32^\circ + 90^\circ = 122^\circ$

Thus,
$\angle UQR = \angle PSR = 122^\circ$

1. Find the interior angles of the original parallelogram PQRS.

In the original parallelogram $PQRS$, we are given in the folding diagram that after folding along the vertical line $TS$, the remaining angle is $32^\circ$.
Since $TS \perp PQ$, in the unfolded state, the angle at $S$ of the right-angled triangle is $90^\circ - 58^\circ = 32^\circ$.
Therefore, the original angle $\angle PSR$ of the parallelogram is:
$\angle PSR = 90^\circ + 32^\circ = 122^\circ$
Since adjacent angles in a parallelogram sum to $180^\circ$, the original angle $\angle QRS$ is:
$\angle QRS_{\text{original}} = 180^\circ - 122^\circ = 58^\circ$

2. Use the symmetry of the fold UR to find the angle $\angle QRU$.

When the corner containing $Q$ is folded along the line $UR$, the original angle $\angle QRS_{\text{original}}$ is divided into three parts: the folded angle $\angle QRU$, the overlapping region underneath (which is also equal to $\angle QRU$ due to folding symmetry), and the remaining unfolded angle $\angle QRS_{\text{remaining}} = 14^\circ$ shown in Figure 2.

Thus, the folding relation gives:
$2 \times \angle QRU + 14^\circ = \angle QRS_{\text{original}}$
$2 \times \angle QRU + 14^\circ = 58^\circ$
$2 \times \angle QRU = 58^\circ - 14^\circ = 44^\circ$
$\angle QRU = 22^\circ$